Construct a parallelogram subject to certain conditions I am having trouble with the following exercise from Dollon and Gilet's Géométrie plane.

Two parallel lines $\Delta$ and $\Delta'$ are given, as well as a point $A$ on $\Delta$ and a point $O$ on neither line.

*

*Construct two parallel lines passing respectively through $A$ and $O$ and forming a rhombus with $\Delta$ and $\Delta'$.


*Deduce from this a parallelogram with given perimeter two of whose sides pass through $O$ and $A$, the other sides lying on $\Delta$ and $\Delta'$, respectively.

I have solved the first part of the problem. (Draw two parallel lines through $O$ and $A$ that are the same distance apart as $\Delta$ and $\Delta'$.) However, I am unable to see how this leads to a solution to the second part of the question.
Edit I should point out that when the problem says that one of the sides must pass through $O$, the most appropriate interpretation is that by “side” the problem means the whole line, not the segment.
Also, I will clarify a point that came up in the comments concerning the meaning of the phrase “with given perimeter.” In addition to the lines $\Delta$ and $\Delta'$ and the points $A$ and $O$, we are given some segment of length $l$ (the “given perimeter”). The problem is to construct from these data, using ruler and compass, a parallelogram whose sides satisfy the stated conditions and whose perimeter is $l$.
Edit Here is a figure.

You are given the lines $\Delta$, $\Delta'$, the points $A$ and $O$, and the length $l$. (To save space, I've drawn $l/2$ instead.) The problem is to construct the slanted lines so that the resulting parallelogram has perimeter $l$.
 A: Okay, this was easier than I thought. I would delete the question, but you're not allowed to delete questions with bounties. 
Construct a point $B$ at a distance $l/2$ from $A$ going right (say). Then apply the first part of the question, but with $B$ playing the role of $A$.
A: Shall I call the parallel lines $AH, RP?$
Let us construct a rectangle of dimensions (vertical,horizontal) $(v,h)$, and skew it by an angle $  \gamma $ so the oblique/distorted  parallelogram has the same perimeter.
$$ 2 h + \frac{2 v}{\sin \gamma }  = l\;;\  \;  \; \sin \gamma = \frac{v}{l/2 -h }\;;  $$
An auxilary semi-circle is sketched by  Ruler and  Compass to construct angle $ \gamma $ to transfer the angle to $H$ next. The diameter $ (l/2 -h )$ is got by permissible subtraction/division of line segments.
Rotate the given line $AO$ about $A$ so that $O$ falls on horizontal line at $H$.This entails rotation $- \alpha $. Draw a line at $\gamma$ transferred to horizontal position as slant $HQ$ and complete the blue parallelogram by drawing a parallel at $A$ which has required perimeter.
Now rotate the parallelogram back as a rigid figure ( so that $v,l$ remain the  same ) by angle $ +\alpha$ to the position indicated  by the red parallelogram finally. 
EDIT1:
This response was the earlier one,now superseded. There was difficulty in figuring out what was required, whether it was interesting? etc.

A: $AB$ is balance length after subtracting $OA$ from given total perimeter length. $OABP$ is a parallelogram but the figure between $ \Delta,\Delta^{\prime}$ is a rhombus as the adjacent sides have equal length as radii.It is now probably  more clear what he says.It is too simple...

